Skip to content

Commit 88eda20

Browse files
author
Administrator
committed
commit
1 parent 3c7c449 commit 88eda20

File tree

91 files changed

+3171
-0
lines changed

Some content is hidden

Large Commits have some content hidden by default. Use the searchbox below for content that may be hidden.

91 files changed

+3171
-0
lines changed

‎.idea/modules.xml

Lines changed: 35 additions & 0 deletions
Some generated files are not rendered by default. Learn more about customizing how changed files appear on GitHub.

‎【054】【SpiralMatrix】/src/Solution.java

Lines changed: 111 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,111 @@
1+
import java.util.ArrayList;
2+
import java.util.List;
3+
4+
/**
5+
* Author: 王俊超
6+
* Date: 2015-08-21
7+
* Time: 17:11
8+
* Declaration: All Rights Reserved !!!
9+
*/
10+
public class Solution {
11+
/**
12+
* <pre>
13+
* 原题
14+
* Given a matrix of m x n elements (m rows, n columns), return all elements
15+
* of the matrix in spiral order.
16+
* For example,
17+
* Given the following matrix:
18+
*
19+
* [
20+
* [ 1, 2, 3 ],
21+
* [ 4, 5, 6 ],
22+
* [ 7, 8, 9 ]
23+
* ]
24+
*
25+
* You should return [1,2,3,6,9,8,7,4,5].
26+
* 题目大意
27+
* 给定一个m*n的矩阵，输入所有元素的螺旋顺序。
28+
*
29+
* 解题思路
30+
* 使用计算输出的方法，先处理上面一行，再处理右边一列，再处理下面一行，再处理左边一列，
31+
* 一直这样操作，直到所有的元素都处理完。
32+
* </pre>
33+
*
34+
* @param matrix
35+
* @return
36+
*/
37+
public List<Integer> spiralOrder(int[][] matrix) {
38+
List<Integer> result = new ArrayList<>(50);
39+
40+
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
41+
return result;
42+
}
43+
44+
// 只有一行的情况
45+
if (matrix.length == 1) {
46+
for (int i : matrix[0]) {
47+
result.add(i);
48+
}
49+
50+
return result;
51+
}
52+
53+
// 只有一列的情况
54+
if (matrix[0].length == 1) {
55+
for (int i = 0; i < matrix.length; i++) {
56+
result.add(matrix[i][0]);
57+
}
58+
59+
return result;
60+
}
61+
62+
// 计算有多少圈
63+
int row = matrix.length;
64+
int col = matrix[0].length;
65+
int cycle = row < col ? row : col;
66+
cycle = (cycle + 1) / 2;
67+
68+
int round = 0; // 记录当前是第几圈
69+
int left = 0;
70+
int right = matrix[0].length - 1;
71+
int top = 0;
72+
int down = matrix.length - 1;
73+
int total = col * row;
74+
int count = 0;
75+
while (round < cycle) {
76+
77+
// 上面一行
78+
for (int i = left; i <= right && count < total; i++) {
79+
count++;
80+
result.add(matrix[round][i]);
81+
}
82+
top++; //
83+
84+
// 右边一列
85+
for (int i = top; i <= down && count < total; i++) {
86+
count++;
87+
result.add(matrix[i][col - round - 1]);
88+
}
89+
right--;
90+
91+
// 底下一行
92+
for (int i = right; i >= left && count < total; i--) {
93+
count++;
94+
result.add(matrix[row - round - 1][i]);
95+
96+
}
97+
down--;
98+
99+
// 左边一列
100+
for (int i = down; i >= top && count < total; i--) {
101+
count++;
102+
result.add(matrix[i][round]);
103+
}
104+
left++;
105+
106+
round++;
107+
}
108+
109+
return result;
110+
}
111+
}

‎【054】【SpiralMatrix】/【054】【SpiralMatrix】.iml

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
<?xml version="1.0" encoding="UTF-8"?>
2+
<module type="JAVA_MODULE" version="4">
3+
<component name="NewModuleRootManager" inherit-compiler-output="true">
4+
<exclude-output />
5+
<content url="file://$MODULE_DIR$">
6+
<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
7+
</content>
8+
<orderEntry type="inheritedJdk" />
9+
<orderEntry type="sourceFolder" forTests="false" />
10+
</component>
11+
</module>

‎【058】【LengthOfLastWord】/src/Solution.java

Lines changed: 54 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,54 @@
1+
/**
2+
* Author: 王俊超
3+
* Date: 2015-08-21
4+
* Time: 19:27
5+
* Declaration: All Rights Reserved !!!
6+
*/
7+
public class Solution {
8+
/**
9+
* <pre>
10+
* 原题
11+
* Given a string s consists of upper/lower-case alphabets and empty space characters ' ',
12+
* return the length of last word in the string.
13+
* If the last word does not exist, return 0.
14+
* Note: A word is defined as a character sequence consists of non-space characters only.
15+
* For example,
16+
* Given s = "Hello World",
17+
* return 5.
18+
*
19+
* 题目大意
20+
* 给定一个由大小写字母组和空格组成的字符串，返回字符串中的最后一个单词长度。
21+
*
22+
* 解题思路
23+
* 先从后找第一个字母的位置x，如果没有找到就返回0，如果找到，再找第一个空格的位记为y(y可能是-1，
24+
* 因为没有找到空格)，返回结果x-y。
25+
* </pre>
26+
*
27+
* @param s
28+
* @return
29+
*/
30+
public int lengthOfLastWord(String s) {
31+
32+
int index = s.length() - 1;
33+
34+
// 从后面向前找第一个不是' '的字符
35+
while (index >= 0 && s.charAt(index) == ' ') {
36+
index--;
37+
}
38+
39+
if (index < 0) {
40+
return 0;
41+
}
42+
43+
int tmp = index;
44+
45+
// 执行到下面说明存在最后一个单词
46+
47+
// 从后面向前找第一个是' '的字符
48+
while (index >= 0 && s.charAt(index) != ' ') {
49+
index--;
50+
}
51+
52+
return tmp - index;
53+
}
54+
}

‎【058】【LengthOfLastWord】/【058】【LengthOfLastWord】.iml

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
<?xml version="1.0" encoding="UTF-8"?>
2+
<module type="JAVA_MODULE" version="4">
3+
<component name="NewModuleRootManager" inherit-compiler-output="true">
4+
<exclude-output />
5+
<content url="file://$MODULE_DIR$">
6+
<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
7+
</content>
8+
<orderEntry type="inheritedJdk" />
9+
<orderEntry type="sourceFolder" forTests="false" />
10+
</component>
11+
</module>

‎【059】【SpiralMatrixII】/src/Solution.java

Lines changed: 75 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,75 @@
1+
/**
2+
* Author: 王俊超
3+
* Date: 2015-08-21
4+
* Time: 19:28
5+
* Declaration: All Rights Reserved !!!
6+
*/
7+
public class Solution {
8+
/**
9+
* <pre>
10+
* 原题
11+
* Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
12+
* For example,
13+
* Given n = 3,
14+
* You should return the following matrix:
15+
*
16+
* [
17+
* [ 1, 2, 3 ],
18+
* [ 8, 9, 4 ],
19+
* [ 7, 6, 5 ]
20+
* ]
21+
*
22+
* 题目大意
23+
* 给定一个整数n，生成一个n*n的矩阵，用1-n^2的数字进行螺旋填充。
24+
*
25+
* 解题思路
26+
* 采用计算生成法，对每一个位置计算对应的数。
27+
* </pre>
28+
*
29+
* @param n
30+
* @return
31+
*/
32+
public int[][] generateMatrix(int n) {
33+
int[][] result = new int[n][n];
34+
35+
int layer;
36+
int k;
37+
for (int i = 0; i < n; i++) {
38+
for (int j = 0; j < n; j++) {
39+
layer = layer(i, j, n); // 当前坐标外有几层
40+
// n * n - layer * layer外围层使用的最后一个数字（也是最大的）
41+
// 坐标所在的当前层使用的第一个数字
42+
k = n * n - (n - 2 * layer) * (n - 2 * layer) + 1;
43+
result[i][j] = k;
44+
45+
// (n - 2 * layer - 1)：四个(n - 2 * layer - 1)就是（x，y）坐标所在层的所有元素个数
46+
if (i == layer) { // 情况一、坐标离上边界最近
47+
result[i][j] = k + j - layer;
48+
} else if (j == n - layer - 1) { // 情况二、坐标离右边界最近
49+
result[i][j] = k + (n - 2 * layer - 1) + i - layer;
50+
} else if (i == n - layer - 1) { // 情况三、坐标离下边界最近
51+
result[i][j] = k + 3 * (n - 2 * layer - 1) - (j - layer);
52+
} else { // 情况三、坐标离左边界最近
53+
result[i][j] = k + 4 * (n - 2 * layer - 1) - (i - layer);
54+
}
55+
}
56+
}
57+
58+
return result;
59+
}
60+
61+
/**
62+
* 在一个n*n的矩阵中，计算（x,y）坐标外有多少层，坐标从0开始计算
63+
*
64+
* @param x 横坐标
65+
* @param y 纵坐标
66+
* @param n 矩阵大小
67+
* @return 坐标外的层数
68+
*/
69+
public int layer(int x, int y, int n) {
70+
x = x < n - 1 - x ? x : n - 1 - x; // 计算横坐标离上下边界的最近距离
71+
y = y < n - 1 - y ? y : n - 1 - y; // 计算纵坐标离左右边界的最近距离
72+
73+
return x < y ? x : y; // 较小的值为坐标的外围层数
74+
}
75+
}

‎【059】【SpiralMatrixII】/【059】【SpiralMatrixII】.iml

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
<?xml version="1.0" encoding="UTF-8"?>
2+
<module type="JAVA_MODULE" version="4">
3+
<component name="NewModuleRootManager" inherit-compiler-output="true">
4+
<exclude-output />
5+
<content url="file://$MODULE_DIR$">
6+
<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
7+
</content>
8+
<orderEntry type="inheritedJdk" />
9+
<orderEntry type="sourceFolder" forTests="false" />
10+
</component>
11+
</module>

‎【062】【UniquePaths】/src/Solution.java

Lines changed: 63 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,63 @@
1+
/**
2+
* Author: 王俊超
3+
* Date: 2015-08-21
4+
* Time: 19:30
5+
* Declaration: All Rights Reserved !!!
6+
*/
7+
public class Solution {
8+
/**
9+
* <pre>
10+
* 原题
11+
* A robot is located at the top-left corner of a m x n grid
12+
* (marked ‘Start’ in the diagram below).
13+
* The robot can only move either down or right at any point in time.
14+
* The robot is trying to reach the bottom-right corner of the grid
15+
* (marked ‘Finish’ in the diagram below).
16+
* How many possible unique paths are there?
17+
*
18+
* Above is a 3 x 7 grid. How many possible unique paths are there?
19+
* Note: m and n will be at most 100.
20+
*
21+
* 题目大意
22+
* 一个机器人在一个m*n的方格的左上角。
23+
* 机器人只能向右或都向下走一个方格，机器人要到达右下角的方格。
24+
* 请问一共有多少种唯一的路径。
25+
* 注意：ｍ和ｎ最大不超100。
26+
*
27+
* 解题思路
28+
* 典型的动态规划问题，对问题使用动态规划的方法进行求解。
29+
* 用一个ｍ*n的组数A保存结果。
30+
* 对于A数组中的元素有。
31+
* 1、当x=0或者y=0时有A[x][y] = 1；
32+
* 2、当x>=1并且y>=1时有A[\x][\y] = A[x-1][y]+A[\x][y-1]。
33+
* 3、所求的结点就是A[m-1][n-1]。
34+
* </pre>
35+
*
36+
* @param m
37+
* @param n
38+
* @return
39+
*/
40+
public int uniquePaths(int m, int n) {
41+
int[][] result = new int[m][n];
42+
43+
// 第一列的解
44+
for (int i = 0; i < m; i++) {
45+
result[i][0] = 1;
46+
}
47+
48+
// 第一行的解
49+
for (int i = 1; i < n; i++) {
50+
result[0][i] = 1;
51+
}
52+
53+
// 其它位置的解
54+
for (int i = 1; i < m; i++) {
55+
for (int j = 1; j < n; j++) {
56+
result[i][j] = result[i - 1][j] + result[i][j - 1];
57+
}
58+
}
59+
60+
// 所求的解
61+
return result[m - 1][n - 1];
62+
}
63+
}

‎【062】【UniquePaths】/【062】【UniquePaths】.iml

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
<?xml version="1.0" encoding="UTF-8"?>
2+
<module type="JAVA_MODULE" version="4">
3+
<component name="NewModuleRootManager" inherit-compiler-output="true">
4+
<exclude-output />
5+
<content url="file://$MODULE_DIR$">
6+
<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
7+
</content>
8+
<orderEntry type="inheritedJdk" />
9+
<orderEntry type="sourceFolder" forTests="false" />
10+
</component>
11+
</module>

0 commit comments

Comments
 (0)